Cantor set exercise This is an exercise from Abbott's real analysis book. It's exercise 3.4.4.(b) on page 93. I couldn't find a definition of ''dimension'' in the book. The only thing I could find is something on page 77. It seems to say that if you "magnify" an object by a factor of $3$ and denote the dimension by $d$ then $3^d = $ size of the magnified object. Example: if the object is a square of length $1$ then scaling it by $3$ is $9$ times the square  $\implies$ $3^d = 9$ $\implies d=2$. I tried to apply this to this exercise:
Let $C$ now be the fat Cantor set: Let $C_0 = [0,1]$, $C_1= [0,3/8] \cup [5/8,1]$ and so on (remove the middle fourth in each step). 
$C = \bigcap_n C_n$ is closed since it is an intersection of closed sets and it is bounded since it is contained in $[0,1]$ therefore $C$ is compact. It is perfect because it is closed and because it does not contain any isolated points: if $x \in C$ then for every $n$, $x \in C_n$ and since the endpoints of $C_n$ are in $C_n$ there is a point other than $x$ in $B(x, 2/8(2/3)^n) \cap C$. 
The length of $C$ is $1$ minus everything that is removed: 
$$ 1- 1/4 -  2/8 \sum_{k=0}^\infty (2/3)^k = 1-1/2 = 1/2$$
Its dimension is computed by starting with $[0,3]$ and noting that removing the middle fourth yields two intervals of length $11/8$. Using these to produce two Cantor sets yields two Cantor sets of length $11/8$ each. Therefore the dimension of the new Cantor set (Cantor set produced from $[0,3]$) is 
$$ 3^d = 2 \times {11 \over 8} = {22 \over 8}$$
solving for the dimension of the Cantor set $d$:
$$ d = {\log {22\over 8}\over \log 3} = {\log 22 - \log 8 \over \log 3}$$
and I don't know how to simplify this expression further. Is this $d$ the dimension of the fat Cantor set?
Edit I founbd this. According to this the Hausdorff dimension should be $1$. So my calculation should be false.
I also found this. According to this, if $N$ is the number of self similar copies and $s$ is the scaling factor then $N= s^d$. I computed that there are ${22\over 8}$ self similar copies. So my calculation should be correct.
 A: Your description

remove the middle fourth in each step

doesn't match the linked construction. That probably explains a lot of the confusion.

The Hausdorff dimension of a subset $S$ of a metric space is defined as
$$\dim_{\mathcal{H}} S := \inf \left\{ d > 0 : \mathcal{H}^d(S) = 0 \right\},$$
where $\mathcal{H}^d$ is the $d$-dimensional Hausdorff measure. The Hausdorff measures have the properties


*

*$\mathcal{H}^d(S) < \infty \land d < \delta \implies \mathcal{H}^\delta(S) = 0$, and

*$\mathcal{H}^d(s) > 0 \land \delta < d\implies \mathcal{H}^\delta(S) = \infty$,


so we have $\mathcal{H}^d(S) = 0$ for all $d > \dim_{\mathcal{H}} S$ and $\mathcal{H}^\delta(S) = \infty$ for all $0 \leqslant \delta < \dim_{\mathcal{H}} S$. The Hausdorff measure in the dimension of $S$, $\mathcal{H}^{\dim_{\mathcal{H}} S}(S)$, can be $0$, $\infty$, or a positive real number.

In the construction of the Smith-Volterra-Cantor set, in the $n$-th step one removes an interval of length $2^{-2n}$ from each of the $2^{n-1}$ intervals of $C_n$, and thus the total (Lebesgue) measure of the removed sets is
$$\sum_{n=1}^\infty \frac{2^{n-1}}{2^{2n}} = \sum_{n=1}^\infty \frac{1}{2^{n+1}} = \frac{1}{2},$$
leaving a fat Cantor set of (Lebesgue) measure $\frac{1}{2}$. Now it is a (not entirely trivial) fact that on $\mathbb{R}^k$ the $k$-dimensional Hausdorff measure is a constant multiple of the Lebesgue measure (both are translation-invariant measures, and one has to verify that the unit cube [or unit ball] has finite nonzero $k$-dimensional Hausdorff measure for the conclusion). Thus the $1$-dimensional Hausdorff measure $\mathcal{H}^1(C)$ is a positive real number, which implies that the Hausdorff-dimension of $C$ is $1$.
This fat Cantor set is not self-similar, since the ratio of the length of the removed middle parts to the length of the containing intervals depends on the step number. In the first step, the ratio is $\frac{4^{-1}}{1} = \frac{1}{4}$, in the second, the ratio is $\frac{4^{-2}}{3/8} = \frac{1}{6}$, in the third, it is $\frac{4^{-3}}{5/32} = \frac{1}{10}$. After the $n$-th step, $C_n$ consists of $2^n$ intervals each of length $\frac{2^n+1}{2^{2n+1}}$, and the ratio of the part removed in the next step to the intervals is $\frac{2^{-(2n+2)}}{(2^n+1)/2^{-(2n+1)}} = \frac{1}{2(2^n+1)}$.
Since the set is not self-similar, the dimension cannot be determined by the self-similarity method. But, since the set has positive Lebesgue measure, its (Hausdorff) dimension can be determined from that.

In the construction where in each step the middle fourth of the intervals is removed, we obtain a self-similar Cantor set of Lebesgue measure $0$. Each $\tilde{C}_{n+1}$ has $\frac{3}{4}$ the Lebesgue measure of $\tilde{C}_n$, since always one fourth of the set is removed, so $\lambda(\tilde{C}_n) = \left(\frac{3}{4}\right)^n$, and hence $\lambda(\tilde{C}) = 0$. But
$$\tilde{C} = \frac{3}{8}\cdot\tilde{C} \cup \left(1 - \frac{3}{8}\cdot\tilde{C}\right),$$
where the union is disjoint (and the parts even have positive distance), so for the $d$-dimensional Hausdorff measure we have
$$\mathcal{H}^d(\tilde{C}) = 2\cdot \left(\frac{3}{8}\right)^d\cdot \mathcal{H}^d(\tilde{C}),$$
and if $\mathcal{H}^d(\tilde{C})$ is a nonzero real number, we have
$$2\cdot \left(\frac{3}{8}\right)^d = 1,$$
and that is equivalent to
$$d = \frac{\log 2}{\log \frac{8}{3}}.$$
The self-similarity theorem tells us that the Hausdorff dimension of $\tilde{C}$ is $d = \frac{\log 2}{\log \frac{8}{3}}$ even without knowing whether $\mathcal{H}^d(\tilde{C})$ is zero, infinite or finite and positive.
